Explanation Given: The function f ( x , y ) = 1 − 2 x 2 − 3 y 2 (1) Result used: Second derivative test “Let D ( x , y ) = f x x ( x , y ) f y y ( x , y ) − f 2 x y ( x , y ) . Then, check the values of D ( a , b ) as below. (a) If D ( a , b ) > 0 and f x x < 0 then the function f has a relative maximum at the critical point ( a , b ) . (b) If D ( a , b ) > 0 and f x x > 0 then the function f has a relative minimum at the critical point ( a , b ) . (c) If D ( a , b ) < 0 then the function f has a no relative extrema at the critical point ( a , b ) . The point ( a , b , f ( a , b ) ) is said to be a saddle point (d) If D ( a , b ) = 0 then there will be no result. It is to be noted that in (a) and (b) f x x and f y y must have the same sign to satisfy the condition D ( a , b ) > 0 .” Calculation: Obtain the critical point of the function . The first derivatives of the given function is computed as follows, Partially differentiate the equation (1) with respect to x. ∂ ∂ x f ( x , y ) = ∂ ∂ x ( 1 − 2 x 2 − 3 y 2 ) ∂ ∂ x f ( x , y ) = − 4 x (2) Partially differentiate the equation (1) with respect to y . ∂ ∂ y f ( x , y ) = ∂ ∂ y ( 1 − 2 x 2 − 3 y 2 ) ∂ ∂ y f ( x , y ) = − 6 y (3) Thus, f x = − 4 x and f y = − 6 y . Substitute f x ( x , y ) = 0 and f y ( x , y ) = 0 . That is, − 4 x = 0 and − 6 y = 0 . Solve the above system of equations. The value of x is obtained as follows, − 4 x = 0 x = 0 Similarly, the value of y is obtained as follows, − 6 y = 0 y = 0 Thus, the critical point of the given function is ( 0 , 0 )

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Chapter 8.3, Problem 1E

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To find: The critical points and relative extrema of the function; classify the nature of each points if possible.

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The critical points and relative extrema of the function; classify the nature of each points if possible.

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